Find $\dfrac{d}{dx}\left[ \sqrt{3\cos^3(x)} \right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{9\cos^2(x)\sin(x)}{2\sqrt{3\cos^3(x)}}$ (Choice B) B $-\dfrac{3\sin(x)}{2\sqrt{3\cos(x)}}$ (Choice C) C $\dfrac{1}{2\sqrt{3\cos^3(x)}}$ (Choice D) D $\dfrac{9\cos^2(x)}{\sqrt{3\cos^3(x)}}$
Explanation: $\sqrt{3\cos^3(x)}$ is a composition of three functions! Let... $u(x)=\sqrt{x}$ $v(x)=3x^3$ $w(x)=\cos(x)$... then $\sqrt{3\cos^3(x)}=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $\dfrac{d}{dx}\left[ \sqrt{3\cos^3(x)} \right]$, we will need to use the chain rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\dfrac{1}{2\sqrt{x}}$ $v'(x)=9x^2$ $w'(x)=-\sin(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={\dfrac{1}{2\sqrt{3(\cos(x))^3}}}\cdot{ 9(\cos(x))^2}\cdot{-\sin(x)} \\\\ &=-\dfrac{9\cos^2(x)\sin(x)}{2\sqrt{3\cos^3(x)}} \end{aligned}$ In conclusion: $\dfrac{d}{dx}\left[ \sqrt{3\cos^3(x)} \right]=-\dfrac{9\cos^2(x)\sin(x)}{2\sqrt{3\cos^3(x)}}$